Solve {l}{x+y/4=9}{2x/5+y=18}

$$\left\{ \begin{array} { l } { x + \frac { y } { 4 } = 9 } \\ { \frac { 2 x } { 5 } + y = 18 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=16$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 4.

$$4x+y=36$$

Consider the second equation. Multiply both sides of the equation by 5.

$$2x+5y=90$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$4x+y=36,2x+5y=90$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$4x+y=36$$

Subtract y from both sides of the equation.

$$4x=-y+36$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(-y+36\right)$$

Multiply \frac{1}{4} times -y+36.

$$x=-\frac{1}{4}y+9$$

Substitute -\frac{y}{4}+9 for x in the other equation, 2x+5y=90.

$$2\left(-\frac{1}{4}y+9\right)+5y=90$$

Multiply 2 times -\frac{y}{4}+9.

$$-\frac{1}{2}y+18+5y=90$$

Add -\frac{y}{2} to 5y.

$$\frac{9}{2}y+18=90$$

Subtract 18 from both sides of the equation.

$$\frac{9}{2}y=72$$

Divide both sides of the equation by \frac{9}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=16$$

Substitute 16 for y in x=-\frac{1}{4}y+9. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{4}\times 16+9$$

Multiply -\frac{1}{4} times 16.

$$x=-4+9$$

Add 9 to -4.

$$x=5$$

The system is now solved.

$$x=5,y=16$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 4.

$$4x+y=36$$

Consider the second equation. Multiply both sides of the equation by 5.

$$2x+5y=90$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$4x+y=36,2x+5y=90$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&1\\2&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36\\90\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&1\\2&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&1\\2&5\end{matrix}\right))\left(\begin{matrix}4&1\\2&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\2&5\end{matrix}\right))\left(\begin{matrix}36\\90\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\2&5\end{matrix}\right))\left(\begin{matrix}36\\90\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\2&5\end{matrix}\right))\left(\begin{matrix}36\\90\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{4\times 5-2}&-\frac{1}{4\times 5-2}\\-\frac{2}{4\times 5-2}&\frac{4}{4\times 5-2}\end{matrix}\right)\left(\begin{matrix}36\\90\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{18}&-\frac{1}{18}\\-\frac{1}{9}&\frac{2}{9}\end{matrix}\right)\left(\begin{matrix}36\\90\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{18}\times 36-\frac{1}{18}\times 90\\-\frac{1}{9}\times 36+\frac{2}{9}\times 90\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\16\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=16$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 4.

$$4x+y=36$$

Consider the second equation. Multiply both sides of the equation by 5.

$$2x+5y=90$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$4x+y=36,2x+5y=90$$

To make 4x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 4.

$$2\times 4x+2y=2\times 36,4\times 2x+4\times 5y=4\times 90$$

Simplify.

$$8x+2y=72,8x+20y=360$$

Subtract 8x+20y=360 from 8x+2y=72 by subtracting like terms on each side of the equal sign.

$$8x-8x+2y-20y=72-360$$

Add 8x to -8x. Terms 8x and -8x cancel out, leaving an equation with only one variable that can be solved.

$$2y-20y=72-360$$

Add 2y to -20y.

$$-18y=72-360$$

Add 72 to -360.

$$-18y=-288$$

Divide both sides by -18.

$$y=16$$

Substitute 16 for y in 2x+5y=90. Because the resulting equation contains only one variable, you can solve for x directly.

$$2x+5\times 16=90$$

Multiply 5 times 16.

$$2x+80=90$$

Subtract 80 from both sides of the equation.

$$2x=10$$

Divide both sides by 2.

$$x=5$$

The system is now solved.

$$x=5,y=16$$